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Q.

The angle of elevation of a jet fighter from a point A on the ground is 60 °  . After a flight of 15 seconds, the angle of elevation changes to 30 °  . If the jet is flying at a speed of 720 km/hr  , find the constant height.

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a

2.59 km

b

3.59 km

c

4.59 km

d

5.59 km 

answer is A.

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Detailed Solution

Given that the angle of elevation of a jet fighter from a point A on the ground is 60 °  .
We have to find the constant height.
The value for the trigonometric identities can be evaluated by using the right-angled triangle the value for tanθ   can be expressed as tanθ=  opposite   adjacent   
Suppose, B and C be the two positions of the jet fighter and the constant height of the jet fighter be BE=CD=h.
Since the speed of jet fighter is 720 km/hr   Thus, the distance covered in 15 seconds is, BC = 720×1000×15 3600 = 720×150 36 =3000 m =3 km   The required figure geometry is shown in the figure below,
Question ImageThen by using the figure geometry, in triangle BCA
tanθ = BE AE tan 60 ° = h AE 3 = h AE AE = h 3   Similarly, in a triangle CDA
tanθ = CD AD tan 30 ° = h AD 1 3 = h AD AD =h 3   AD =AE+ED 3 h = h 3 +3 2h =3 3 h =2.59 km  
The constant height is 2.59 Km.
Therefore, the correct option is 1.
 
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